<Tri-2024-2168>, hint

Suppose PAB is fixed and C moves on circle of radius 3 (=PC) , then

area of ABC is max when PC is perpendicular to AB. Thus area of ABC is max 

when P is orthocenter(=H) of ABC. In triangle HBC (HB=2, HC=3, <BHC=180-60(<A)=120

[ABC]=1/2 2x3 sin 120=(3/2) root(3). by cosine 2nd law BC=root(19)

Hence altitude of HBC  =[HBC]/BC={(3/2)root(3)}/{root(19)}

Altitude of ABC={(3/2)root(3)}/{root(19)}+1

Max area of ABC =0.5 x root(19) x<{(3/2)root(3)}/{root(19)}+1>